If  3x+4y=122 is a tangent to the ellipse  x2a2+y29=1 for some a∈R, then the distance between the foci the of ellipse is

3x+4y=122⇒4y=−3x+122⇒y=−34x+32Condition of tangency c2=a2m2+b2⇒18=a2.916+9⇒a2.916=9⇒a2=16⇒a=4⇒e=1−b2a2=1−916=74∴ae=74.4=7 ⇒foci are   ±7,0∴ distance between foci =27